How to determine the sign in the interval method. The interval method: solving the simplest strict inequalities

Spacing Method is a simple way to solve the fractional rational inequalities. This is the name of inequalities containing rational (or fractional-rational) expressions that depend on a variable.

1. Consider, for example, the following inequality

The interval method allows you to solve it in a couple of minutes.

On the left side of this inequality is a fractional rational function. Rational, because it contains neither roots, nor sines, nor logarithms - only rational expressions. On the right is zero.

The interval method is based on the following property of a fractional rational function.

A fractional rational function can change sign only at those points where it is equal to zero or does not exist.

Recall how to factorize square trinomial, that is, an expression of the form .

Where and are the roots quadratic equation.

We draw an axis and arrange the points at which the numerator and denominator vanish.

The zeros of the denominator and are punctured points, since at these points the function on the left side of the inequality is not defined (you cannot divide by zero). The zeros of the numerator and - are shaded because the inequality is not strict. For and our inequality is satisfied, since both its parts are equal to zero.

These points break the axis into intervals.

Let us determine the sign of the fractional-rational function on the left side of our inequality on each of these intervals. We remember that a fractional rational function can change sign only at those points where it is equal to zero or does not exist. This means that on each of the intervals between the points where the numerator or denominator vanishes, the sign of the expression on the left side of the inequality will be constant - either "plus" or "minus".

And therefore, to determine the sign of the function on each such interval, we take any point belonging to this interval. The one that suits us.
. Take, for example, and check the sign of the expression on the left side of the inequality. Each of the "brackets" is negative. The left side has a sign.

Next interval: . Let's check the sign for . We get that the left side has changed sign to .

Let's take . When the expression is positive - therefore, it is positive on the entire interval from to .

For , the left side of the inequality is negative.

And finally class="tex" alt="(!LANG:x>7"> . Подставим и проверим знак выражения в левой части неравенства. Каждая "скобочка" положительна. Следовательно, левая часть имеет знак .!}

We have found on what intervals the expression is positive. It remains to write the answer:

Answer: .

Please note: the signs on the intervals alternate. This happened because when passing through each point, exactly one of the linear factors changed sign, and the rest kept it unchanged.

We see that the interval method is very simple. To solve a fractional-rational inequality by the method of intervals, we bring it to the form:

Or class="tex" alt="(!LANG:\genfrac()()()(0)(\displaystyle P\left(x \right))(\displaystyle Q\left(x \right)) > 0"> !}, or or .

(on the left side - a fractional-rational function, on the right side - zero).

Then - we mark on the number line the points at which the numerator or denominator vanishes.
These points divide the entire number line into intervals, on each of which the fractional-rational function retains its sign.
It remains only to find out its sign on each interval.
We do this by checking the sign of the expression at any point within the given interval. After that, we write down the answer. That's all.

But the question arises: do the signs always alternate? No not always! We must be careful not to place signs mechanically and thoughtlessly.

2. Let's look at another inequality.

Class="tex" alt="(!LANG:\genfrac()()()(0)(\displaystyle \left(x-2 \right)^2)(\displaystyle \left(x-1 \right) \left(x-3\right))>0"> !}

We again place points on the axis. The points and are punctured because they are the zeros of the denominator. The dot is also punctured, since the inequality is strict.

When the numerator is positive, both factors in the denominator are negative. This is easy to check by taking any number from a given interval, for example, . The left side has the sign:

When the numerator is positive; the first factor in the denominator is positive, the second factor is negative. The left side has the sign:

When the situation is the same! The numerator is positive, the first factor in the denominator is positive, the second is negative. The left side has the sign:

Finally, with class="tex" alt="(!LANG:x>3"> все множители положительны, и левая часть имеет знак :!}

Answer: .

Why was the alternation of characters broken? Because when passing through the point, the multiplier "responsible" for it did not change sign. Consequently, the entire left-hand side of our inequality did not change sign either.

Conclusion: if the linear factor is in an even power (for example, in a square), then when passing through a point, the sign of the expression on the left side does not change. In the case of an odd degree, the sign, of course, changes.

3. Consider more difficult case. It differs from the previous one in that the inequality is not strict:

The left side is the same as in the previous problem. The picture of signs will be the same:

Maybe the answer will be the same? Not! The solution is added This is because at , both the left and right sides of the inequality are equal to zero - therefore, this point is a solution.

Answer: .

In the problem on the exam in mathematics, this situation is often encountered. Here, applicants fall into a trap and lose points. Be careful!

4. What if the numerator or denominator cannot be factored into linear factors? Consider this inequality:

The square trinomial cannot be factorized: the discriminant is negative, there are no roots. But this is good! This means that the sign of the expression is the same for all, and specifically, it is positive. You can read more about this in the article on the properties of a quadratic function.

And now we can divide both sides of our inequality by a value that is positive for all . We arrive at an equivalent inequality:

Which is easily solved by the interval method.

Pay attention - we divided both sides of the inequality by the value, which we knew for sure that it was positive. Of course, in the general case, you should not multiply or divide an inequality by a variable whose sign is unknown.

5 . Consider another inequality, seemingly quite simple:

So I want to multiply it by . But we are already smart, and we will not do this. After all, it can be both positive and negative. And we know that if both parts of the inequality are multiplied by a negative value, the sign of the inequality changes.

We will act differently - we will collect everything in one part and bring it to a common denominator. Zero will remain on the right side:

Class="tex" alt="(!LANG:\genfrac()()()(0)(\displaystyle x-2)(\displaystyle x)>0"> !}

And after that - applicable interval method.

In this lesson, we will continue to solve rational inequalities using the interval method for more complex inequalities. Consider the solution of linear-fractional and quadratic-fractional inequalities and related problems.

Now back to inequality

Let's consider some related tasks.

To find smallest solution inequalities.

Find the number of natural solutions to the inequality

Find the length of the intervals that make up the set of solutions to the inequality.

2. Portal Natural Sciences ().

3. Electronic educational and methodological complex for preparing grades 10-11 for entrance exams in computer science, mathematics, Russian language ().

5. Education Center "Technology of Education" ().

6. College.ru section on mathematics ().

1. Mordkovich A.G. et al. Algebra Grade 9: Taskbook for students of educational institutions / A. G. Mordkovich, T. N. Mishustina et al. - 4th ed. - M .: Mnemosyne, 2002.-143 p.: ill. No. 28 (b, c); 29(b,c); 35(a,b); 37(b,c); 38(a).

First, some lyrics to get a feel for the problem that the interval method solves. Suppose we need to solve the following inequality:

(x − 5)(x + 3) > 0

What are the options? The first thing that comes to mind for most students is the rules "plus times plus makes plus" and "minus times minus makes plus." Therefore, it suffices to consider the case when both brackets are positive: x − 5 > 0 and x + 3 > 0. Then we also consider the case when both brackets are negative: x − 5< 0 и x + 3 < 0. Таким образом, наше неравенство свелось к совокупности двух систем, которая, впрочем, легко решается:

More advanced students will remember (perhaps) that on the left is a quadratic function whose graph is a parabola. Moreover, this parabola intersects the OX axis at the points x = 5 and x = −3. For further work you need to open the brackets. We have:

x 2 − 2x − 15 > 0

Now it is clear that the branches of the parabola are directed upwards, because coefficient a = 1 > 0. Let's try to draw a diagram of this parabola:

The function is greater than zero where it passes above the OX axis. In our case, these are the intervals (−∞ −3) and (5; +∞) - this is the answer.

Please note that the picture shows exactly function diagram, not her schedule. Because for a real graph, you need to calculate coordinates, calculate offsets and other crap, which we don’t need at all now.

Why are these methods ineffective?

So, we have considered two solutions to the same inequality. Both of them turned out to be very cumbersome. The first decision arises - just think about it! is a set of systems of inequalities. The second solution is also not very easy: you need to remember the parabola graph and a bunch of other small facts.

It was a very simple inequality. It has only 2 multipliers. Now imagine that there will be not 2 multipliers, but at least 4. For example:

(x − 7)(x − 1)(x + 4)(x + 9)< 0

How to solve such inequality? Go through all possible combinations of pros and cons? Yes, we will fall asleep faster than we find a solution. Drawing a graph is also not an option, since it is not clear how such a function behaves on the coordinate plane.

For such inequalities, a special solution algorithm is needed, which we will consider today.

What is the interval method

The interval method is a special algorithm designed to solve complex inequalities of the form f (x) > 0 and f (x)< 0. Алгоритм состоит из 4 шагов:

1. Solve the equation f (x) \u003d 0. Thus, instead of an inequality, we get an equation that is much easier to solve;
2. Mark all the obtained roots on the coordinate line. Thus, the straight line will be divided into several intervals;
3. Find out the sign (plus or minus) of the function f (x) on the rightmost interval. To do this, it is enough to substitute in f (x) any number that will be to the right of all the marked roots;
4. Mark marks on other intervals. To do this, it is enough to remember that when passing through each root, the sign changes.

That's all! After that, it remains only to write out the intervals that interest us. They are marked with a “+” sign if the inequality was of the form f (x) > 0, or a “−” sign if the inequality was of the form f (x)< 0.

At first glance, it may seem that the interval method is some kind of tin. But in practice, everything will be very simple. It takes a little practice - and everything will become clear. Take a look at the examples and see for yourself:

Task. Solve the inequality:

(x − 2)(x + 7)< 0

We work on the method of intervals. Step 1: Replace the inequality with an equation and solve it:

(x − 2)(x + 7) = 0

The product is equal to zero if and only if at least one of the factors zero:

x − 2 = 0 ⇒ x = 2;
x + 7 = 0 ⇒ x = −7.

Got two roots. Go to step 2: mark these roots on the coordinate line. We have:

Now step 3: we find the sign of the function on the rightmost interval (to the right of the marked point x = 2). To do this, take any number that more number x = 2. For example, let's take x = 3 (but no one forbids taking x = 4, x = 10 and even x = 10,000). We get:

f(x) = (x − 2)(x + 7);
x=3;
f (3) = (3 − 2)(3 + 7) = 1 10 = 10;

We get that f (3) = 10 > 0, so we put a plus sign in the rightmost interval.

We pass to the last point - it is necessary to note the signs on the remaining intervals. Remember that when passing through each root, the sign must change. For example, to the right of the root x = 2 there is a plus (we made sure of this in the previous step), so there must be a minus on the left.

This minus extends to the entire interval (−7; 2), so there is a minus to the right of the root x = −7. Therefore, there is a plus to the left of the root x = −7. It remains to mark these signs on the coordinate axis. We have:

Let's return to the original inequality, which looked like:

(x − 2)(x + 7)< 0

So the function should be less than zero. This means that we are interested in the minus sign, which occurs only on one interval: (−7; 2). This will be the answer.

Task. Solve the inequality:

(x + 9)(x − 3)(1 − x )< 0

Step 1: Equate the left side to zero:

(x + 9)(x − 3)(1 − x ) = 0;
x + 9 = 0 ⇒ x = −9;
x − 3 = 0 ⇒ x = 3;
1 − x = 0 ⇒ x = 1.

Remember: the product is zero when at least one of the factors is zero. That is why we have the right to equate to zero each individual bracket.

Step 2: mark all the roots on the coordinate line:

Step 3: find out the sign of the rightmost gap. We take any number that is greater than x = 1. For example, we can take x = 10. We have:

f (x) \u003d (x + 9) (x - 3) (1 - x);
x=10;
f (10) = (10 + 9)(10 − 3)(1 − 10) = 19 7 (−9) = − 1197;
f(10) = -1197< 0.

Step 4: Place the rest of the signs. Remember that when passing through each root, the sign changes. As a result, our picture will look like this:

That's all. It remains only to write the answer. Take another look at the original inequality:

(x + 9)(x − 3)(1 − x )< 0

This is an inequality of the form f (x)< 0, т.е. нас интересуют интервалы, отмеченные знаком минус. А именно:

x ∈ (−9; 1) ∪ (3; +∞)

This is the answer.

A note about function signs

Practice shows that the greatest difficulties in the interval method arise at the last two steps, i.e. when placing signs. Many students begin to get confused: what numbers to take and where to put signs.

To finally understand the interval method, consider two remarks on which it is built:

1. A continuous function changes sign only at the points where it is equal to zero. Such points break the coordinate axis into pieces, within which the sign of the function never changes. That's why we solve the equation f (x) \u003d 0 and mark the found roots on a straight line. The numbers found are the "boundary" points separating the pluses from the minuses.
2. To find out the sign of a function on any interval, it is enough to substitute any number from this interval into the function. For example, for the interval (−5; 6) we can take x = −4, x = 0, x = 4 and even x = 1.29374 if we want. Why is it important? Yes, because many students begin to gnaw doubts. Like, what if for x = −4 we get a plus, and for x = 0 we get a minus? Nothing like that will ever happen. All points in the same interval give the same sign. Remember this.

That's all you need to know about the interval method. Of course, we took it apart in the most simple version. There are more complex inequalities - non-strict, fractional and with repeated roots. For them, you can also apply the interval method, but this is a topic for a separate large lesson.

Now I would like to analyze an advanced trick that drastically simplifies the interval method. More precisely, the simplification affects only the third step - the calculation of the sign on the rightmost piece of the line. For some reason, this technique is not held in schools (at least no one explained this to me). But in vain - in fact, this algorithm is very simple.

So, the function sign on the right piece numerical axis. This piece has the form (a; +∞), where a is the largest root of the equation f (x) = 0. In order not to blow our brains, consider a specific example:

(x − 1)(2 + x )(7 − x )< 0;
f (x) \u003d (x - 1) (2 + x) (7 - x);
(x − 1)(2 + x )(7 − x ) = 0;
x − 1 = 0 ⇒ x = 1;
2 + x = 0 ⇒ x = −2;
7 − x = 0 ⇒ x = 7;

We got 3 roots. We list them in ascending order: x = −2, x = 1 and x = 7. Obviously, the largest root is x = 7.

For those who find it easier to reason graphically, I will mark these roots on the coordinate line. Let's see what happens:

It is required to find the sign of the function f (x) on the rightmost interval, i.e. on (7; +∞). But as we have already noted, to determine the sign, you can take any number from this interval. For example, you can take x = 8, x = 150, etc. And now - the same technique that is not taught in schools: let's take infinity as a number. More precisely, plus infinity, i.e. +∞.

"Are you stoned? How can you substitute infinity into a function? perhaps, you ask. But think about it: we do not need the value of the function itself, we only need the sign. Therefore, for example, the values ​​f (x) = −1 and f (x) = −938 740 576 215 mean the same thing: a function on this interval negative. Therefore, all that is required of you is to find the sign that occurs at infinity, and not the value of the function.

In fact, substituting infinity is very simple. Let's go back to our function:

f(x) = (x − 1)(2 + x)(7 − x)

Imagine that x is very big number. A billion or even a trillion. Now let's see what happens in each parenthesis.

First bracket: (x − 1). What happens if you subtract one from a billion? The result will be a number not much different from a billion, and this number will be positive. Similarly with the second bracket: (2 + x). If we add a billion to a deuce, we get a billion with kopecks - this is positive number. Finally, the third bracket: (7 − x ). Here there will be minus a billion, from which a miserable piece in the form of a seven has been “gnawed off”. Those. the resulting number will not differ much from minus a billion - it will be negative.

It remains to find the sign of the whole work. Since we had a plus in the first brackets, and a minus in the last bracket, we get the following construction:

(+) · (+) · (−) = (−)

The final sign is minus! It doesn't matter what the value of the function itself is. The main thing is that this value is negative, i.e. on the rightmost interval there is a minus sign. It remains to complete the fourth step of the interval method: arrange all the signs. We have:

The original inequality looked like:

(x − 1)(2 + x )(7 − x )< 0

Therefore, we are interested in the intervals marked with a minus sign. We write out the answer:

x ∈ (−2; 1) ∪ (7; +∞)

That's the whole trick that I wanted to tell. In conclusion, there is one more inequality, which is solved by the interval method using infinity. To visually shorten the solution, I will not write step numbers and detailed comments. I will write only what really needs to be written when solving real problems:

Task. Solve the inequality:

x (2x + 8)(x − 3) > 0

We replace the inequality with an equation and solve it:

x (2x + 8)(x − 3) = 0;
x = 0;
2x + 8 = 0 ⇒ x = −4;
x − 3 = 0 ⇒ x = 3.

We mark all three roots on the coordinate line (immediately with signs):

There is a plus on the right side of the coordinate axis, because the function looks like:

f(x) = x(2x + 8)(x − 3)

And if we substitute infinity (for example, a billion), we get three positive brackets. Since the original expression must be greater than zero, we are only interested in pluses. It remains to write the answer:

x ∈ (−4; 0) ∪ (3; +∞)

Spacing Method is a special algorithm designed to solve complex inequalities of the form f(x) > 0. The algorithm consists of 5 steps:

1. Solve the equation f(x) = 0. Thus, instead of an inequality, we get an equation that is much easier to solve;
2. Mark all the obtained roots on the coordinate line. Thus, the straight line will be divided into several intervals;
3. Find the multiplicity of the roots. If the roots are of even multiplicity, then we draw a loop over the root. (The root is considered a multiple if there is an even number of identical solutions)
4. Find out the sign (plus or minus) of the function f(x) on the rightmost interval. To do this, it is enough to substitute in f (x) any number that will be to the right of all the marked roots;
5. Mark the signs on the remaining intervals, alternating them.

After that, it remains only to write out the intervals that interest us. They are marked with a “+” sign if the inequality was of the form f(x) > 0, or a “−” sign if the inequality was of the form f(x)< 0.

In the case of non-strict inequalities (≤ , ≥), it is necessary to include in the intervals the points that are the solution of the equation f(x) = 0;

Example 1:

Solve the inequality:

(x - 2)(x + 7)< 0

We work on the method of intervals.

Step 1: replace the inequality with an equation and solve it:

(x - 2)(x + 7) = 0

The product is equal to zero if and only if at least one of the factors is equal to zero:

x - 2 = 0 => x = 2

x + 7 = 0 => x = -7

Got two roots.

Step 2: mark these roots on the coordinate line. We have:

Step 3: we find the sign of the function on the rightmost interval (to the right of the marked point x = 2). To do this, you need to take any number that is greater than the number x = 2. For example, let's take x = 3 (but no one forbids taking x = 4, x = 10, and even x = 10,000).

f(x) = (x - 2)(x + 7)

f(3)=(3 - 2)(3 + 7) = 1*10 = 10

We get that f(3) = 10 > 0 (10 is a positive number), so we put a plus sign in the rightmost interval.

Step 4: you need to mark the signs on the remaining intervals. Remember that when passing through each root, the sign must change. For example, to the right of the root x = 2 there is a plus (we made sure of this in the previous step), so there must be a minus on the left. This minus extends to the entire interval (−7; 2), so there is a minus to the right of the root x = −7. Therefore, there is a plus to the left of the root x = −7. It remains to mark these signs on the coordinate axis.

Let's return to the original inequality, which looked like:

(x - 2)(x + 7)< 0

So the function must be less than zero. This means that we are interested in the minus sign, which occurs only on one interval: (−7; 2). This will be the answer.

Example 2:

Solve the inequality:

(9x 2 - 6x + 1)(x - 2) ≥ 0

Decision:

First you need to find the roots of the equation

(9x 2 - 6x + 1)(x - 2) = 0

Let's collapse the first bracket, we get:

(3x - 1) 2 (x - 2) = 0

x - 2 = 0; (3x - 1) 2 = 0

By solving these equations we get:

Let's plot the points on the number line:

Because x 2 and x 3 are multiple roots, then there will be one point on the line and above it “ the loop”.

Take any number less than the leftmost point and substitute it into the original inequality. Let's take the number -1.

Do not forget to include the solution of the equation (found by X), because our inequality is not strict.

Answer: ()U)